On the planet of Mercury, 4-year-olds average 2.8 hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.6 hours and the amount of time spent alone is normally distributed. We randomly survey one Mercurian 4-year-old living in a rural area. We are interested in the amount of time $X$ the child spends alone per day. (Source: San Jose Mercury News) Round all answers to 4 decimal places where possible.
a. What is the distribution of $\mathrm{X}$ ? $\mathrm{X}-\mathrm{N}\left(\begin{array}{lll}2.8 & \checkmark\end{array}, \begin{array}{lll}1.6 & \checkmark\end{array}\right) 0^{8} 0^{8}$
b. Find the probability that the child spends less than 0.9 hours per day unsupervised.
c. What percent of the children spend over 1.9 hours per day unsupervised. \% (Round to 2 decimal places)
d. $83 \%$ of all children spend at least how many hours per day unsupervised? hours.

Step 1 ：Given that the distribution of X is \(N(2.8, 1.6^2)\), this means that the amount of time spent alone is normally distributed with a mean of 2.8 hours and a standard deviation of 1.6 hours.

Step 2 ：To find the probability that the child spends less than 0.9 hours per day unsupervised, we need to standardize this value and look it up in the standard normal distribution table. The standardized value (z-score) is calculated as follows: \(Z = \frac{X - \mu}{\sigma} = \frac{0.9 - 2.8}{1.6} = -1.1875\)

Step 3 ：Looking up this value in the standard normal distribution table, we find that the probability is approximately 0.1175. So, the probability that the child spends less than 0.9 hours per day unsupervised is approximately \(\boxed{0.1175}\) or 11.75%.

Step 4 ：To find the percent of children who spend over 1.9 hours per day unsupervised, we need to standardize this value and look it up in the standard normal distribution table. The standardized value (z-score) is calculated as follows: \(Z = \frac{X - \mu}{\sigma} = \frac{1.9 - 2.8}{1.6} = -0.5625\)

Step 5 ：Looking up this value in the standard normal distribution table, we find that the probability is approximately 0.2872. However, since we are interested in the percent of children who spend over 1.9 hours per day unsupervised, we need to subtract this value from 1. \(1 - 0.2872 = 0.7128\)

Step 6 ：So, approximately \(\boxed{71.28}\)% of children spend over 1.9 hours per day unsupervised.

Step 7 ：To find the number of hours that 83% of all children spend at least unsupervised, we need to find the z-score that corresponds to 0.83 in the standard normal distribution table. This z-score is approximately 0.954. We then use this z-score to find the corresponding X value: \(X = \mu + Z\sigma = 2.8 + 0.954(1.6) = 4.3264\)

Step 8 ：So, 83% of all children spend at least \(\boxed{4.3264}\) hours per day unsupervised.